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Consider the heat equation $$ u_t - div[a(x,t) \nabla u] =0,\quad (x,t) \in B(r) \times [-r^2, 0] \subset \mathbb R^{d+1} $$ for a Hölder continuous coefficient $a(x,t)$ satisfying $$ 0<C_o \le a(x,t) \le C_1. $$

Now $u$ satisfies the Harnack inequality $$ \sup_{Q^-(r^2,r)} u \le H_o \inf_{Q^+(r^2,r)}u $$ for a constant $H_o = H_o(d, C_o, C_1)>0$, and $$ Q^-(r^2,r) :=B(r/2) \times \left[-\frac{3r^2}4, -\frac{r^2}2\right]\quad and\quad Q^+(r^2,r) :=B(r/2) \times \left[-\frac{r^2}4, 0\right]. $$

My question is that what is the dependence of $H_o$ on $C_o$ and $C_1$. In particular, I am interested in knowing what happens to $H_o$ if $C_o$ and $C_1$ are replaced by $AC_o$ and $AC_1$ for $A\ge 1$. This can, of course, be answered by just going through the proof and checking what happens, but this is quite tedious and I was wondering if there is a simpler way to see this and/or if someone knows this by heart.

Another way to put this is to consider the equation with a scaled time $s=t/A$, which cancels the constant $A$ from the structure conditions, but then again, this also requires scaling the sets in the Harnack inequality, so it does not seem completely obvious what happens to the constant.

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I think that $H_0$ grows like $e^{A}$, which is optimal in light of the example $u(x,t) = e^{-At}\cos(x)$ which solves $u_t - A\Delta u = 0$.

Say we fix $C_i$ and the corresponding Harnack constant is $H$, and that $u$ is positive in $B_2 \times [-4,0]$. The proof is by applying the Harnack inequality to the rescaling you suggested, $\tilde{u}(x,t) = u(x, t/A)$, which is defined on $B_2 \times [-4A,0]$. Iterating the Harnack inequality we have that $\tilde{u}|_{B_1 \times \{-A\}} < H\tilde{u}(0,-A + 1) < H^2 \tilde{u}(0,-A + 2) ... < H^A\tilde{u}(0,0)$. Scaling back gives the Harnack inequality $u|_{B_1 \times \{-1\}} \leq H^Au(0,0)$.

Remark: I think the dependence on the ellipticity ratio $\Lambda = C_1/C_0$ (say we fix $C_0 = 1$ and let $C_1$ get large) can be understood similarly, with dependence going exponentially in $\sqrt{\Lambda}$ as suggested by the stationary example $u = e^{\sqrt{\Lambda}x}\cos(y)$, which solves the elliptic equation $tr(A \cdot D^2u) = 0$ where $A = \text{diag}(1,\Lambda)$. This is because $u$ is harmonic in an ``ellipsoidal geometry'' where the balls are ellipsoids with vertical axis $1$ and horizontal axis $1/\sqrt{\Lambda}$, and the proof by rescaling/iterating in overlapping ellipsoids for constant coefficients works. This is intuitively the worst case scenario, since if the coefficients jump around then the ellipsoids "rotate" and one iterates the Harnack inequality fewer times, but to do things rigorously I think one needs to go through the original proof. (Maybe there is a good probabilistic interpretation of this too.)

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  • $\begingroup$ I thought about the iteration argument, but I was wondering if it possible to do better than that in terms of the constant. But this explicit example you gave of course solves the question. $\endgroup$ – Juhana Siljander Aug 19 '15 at 14:45
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I am also interested in the dependency of the constant in the Harnack inequality. Dependency of the constant in the Harnack inequality

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